# Can we find subgroup of $(\mathbb{R},+)$ with order 2?

We used the following idea: first get a set of Hamel basis for $\mathbb{R}$, secondly, divide it into two parts such that one set of the Hamel basis forms a group, the other one is just the former one with a representative element added. However, it's hard to prove the existence of the representative and the construction of the division is also ambigious.

Thanks for your attention.

-
Every non-zero element of $\Bbb R$ generates an infinite subgroup; the only finite subgroup is the trivial one. – Brian M. Scott Mar 21 '13 at 16:51
$2x=0\iff x=0$. – anon Mar 21 '13 at 17:04
Anon's comment is, in fact, the complete answer, yet I wonder why would you want to mess with Hamel basis and stuff...? – DonAntonio Mar 21 '13 at 17:10
@DonAntonio Messing with Hamel bases and the description makes it seem like maybe the aim is actually to find a subgroup of index $2$ rather than order $2$. I'll leave it up to the OP to clarify if this is the case though. – Tobias Kildetoft Mar 21 '13 at 17:16
Do you mean 'index' instead of 'order'? Also why would you want to deal with division? – Simon Markett Mar 21 '13 at 17:18

## 1 Answer

No. There are no nontrivial elements in $\mathbb{R}$ with finite order.

If you're looking for a subgroup of index 2, you're not going to have much luck either. If $H$ has index $2$ in $\mathbb{R}$, then $\mathbb{R}/H$ is a group of order $2$, so for any $r\in \mathbb{R}$, $2(rH)=(2r)H=H$. But then we can simply take $r=s/2$ given any $s\in \mathbb{R}$ and conclude that $sH=H$. Thus we have $\mathbb{R}\subseteq H$, a contradiction.

If you're into generalizations, by the way, note that we may easily adapt the above to prove that there are no subgroups of finite index in $\mathbb{R}$ at all (or, more generally, in any the additive group of any field of characteristic $0$). (Even more generally, fixing an $n\in \mathbb{N}$, no group with the property that $\forall g\in G \exists h\in G : h^n=g$ may have a subgroup of index $n$; in particular, no ring in which $n\in \mathbb{N}$ is a unit may have an additive subgroup of index $n$.)

-
Any divisible group cannot have proper subgroups of finite index. Now what are these groups? Well, (abelian) groups in which you can divide! Formally, these are groups $G$ in which for every positive integer $n$ and every $g \in G$, there exists an $x \in G$ such that $nx = g$. – Nicky Hekster Mar 21 '13 at 20:40